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Search: id:A154139
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| A154139 |
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Indices k such that 4 plus the k-th triangular number is a perfect square. |
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+0
4
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| 0, 6, 9, 39, 56, 230, 329, 1343, 1920, 7830, 11193, 45639, 65240, 266006, 380249, 1550399, 2216256, 9036390
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 4. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Nov 10 2009]
a(1..4)=(0,6,9,39); a(n>4)=6*a(n-2)-a(n-4)+2. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Nov 10 2009]
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LINKS
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F. T. Adams-Watters, SeqFan Discussion, Oct 2009
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FORMULA
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{k: 4+k*(k+1)/2 in A000290}
Conjecture: a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
Conjecture: G.f.: x^2*(-6-3*x+6*x^2+x^3)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)) = 1+1/2*(4+11*x)/(x^2-2*x-1)+1/2/(x-1)+1/2*(-3+2*x)/(x^2+2*x-1).
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EXAMPLE
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0*(0+1)/2+4 = 2^2. 6*(6+1)/2+4 = 5^2. 9*(9+1)/2+4 = 7^2. 39*(39+1)/2+4 = 28^2.
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MAPLE
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a := proc (n) if type(sqrt(4+(1/2)*n*(n+1)), integer) = true then n else end if end proc: seq(a(n), n = 0 .. 10^7); [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 31 2009]
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CROSSREFS
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Cf. A000217, A000290, A006451.
Sequence in context: A038263 A004989 A147355 this_sequence A126110 A098662 A056425
Adjacent sequences: A154136 A154137 A154138 this_sequence A154140 A154141 A154142
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KEYWORD
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nonn,more,new
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AUTHOR
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R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 18 2009
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EXTENSIONS
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a(17),a(18) from Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 31 2009
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